Question

The value of #((x-y)/z+(y-z)/x+(z-x)/y)(z/(x-y)+x/(y-z)+y/(z-x))# is ? when #x+y+z=0#

Answer 1

The Numerator (NR.) of the Expression (Exp.) is,
The NR.=(x-y)/z+(y-z)/x+(z-x)/y,
={xy(x-y)+yz(y-z)+zx(z-x)} /(xyz),
={x2y-xy2+y2z-yz2+z2x-zx2}/(xyz),
={x2y-zx2-xy2+xz2+y2z-yz2}/(xyz),
={x2(y-z)-x(y2-z2)+yz(y-z)}/(xyz),
={ x2(y-z)-x(y-z)(y+z) +yz(y-z)}/(xyz),
=[(y-z){x2-x(y+z)+yz}]/(xyz),
=[(y-z){x2-xy-xz+yz}]/(xyz),
=[(y-z){x(x-y)-z(x-y)}]/(xyz),
Hence, the NR.={-(y-z)(x-y)(z-x)}/(xyz).

Similarly, the Denominator (DR.) of the Exp. can be shown,
DR=-{z(z-x)(z-y)+y(y-x)(y-z)+x(x-y)(x-z)}/{(y-z)(x-y)(z-x)},
=-{z(z2-(x+y)z+xy)+y(y2-(z+x)y+xz)+x(x2-(y+z)x+yz)}/{(y-z)(x-y)(z-x)}.
Here, from x+y+z=0, we get, x+y=-z, y+z=-x, z+x=-y.
Hence, z2-(x+y)z+xy=z2+z2+xy=2z2+xy, y2-(y+z)x+zx=2y2+zx, x2-(y+z)x+yz=2x2+yz.
Therefore, DR.={-z(2z2+xy)-y(2y2+zx)-z(2x2+yz)}/{(y-z)(x-y)(z-x)},
=-{2z3+2y3+2x3+3xyz}/{(y-z)(x-y)(z-x)}.
Now, recall that, when x+y+z=0, x3+y3+z3=3xyz.
Hence, DR.=-{2(3xyz)+3xyz}/{(y-z)(x-y)(z-x)}.
Thus, DR.=-(9xyz)/{(y-z)(x-y)(z-x)}.
Therefore, the required value=(NR.)/(DR.)=9.

Answer 2

woah

Answer 3

@ratnakermehta20 very nice work !!!